Two equal charges q are kept fixed at `-a and +a` along the x-axis. A particle of mass m and charge `q/2` is brought to the origin and given a small displacement along the x-axis, then

Options

A. the particle executes oscillatory motion
B. the particle remains stationary
C. the particle executes SHM along x-axis
D. the particle executes SHM along y-axis

Explanation

To solve the problem, we need to analyze the forces acting on the charge \( \frac{q}{2} \) when it is displaced from the origin along the x-axis. Here’s the step-by-step solution:

### Step 1: Understand the Setup
We have two equal charges \( q \) located at positions \( -a \) and \( +a \) on the x-axis. A third charge \( \frac{q}{2} \) is placed at the origin (0,0).

### Step 2: Displacement of the Charge
When the charge \( \frac{q}{2} \) is given a small displacement \( x \) along the x-axis, its new position becomes \( x \) (where \( x \) is a small positive or negative value).

### Step 3: Calculate the Forces Acting on \( \frac{q}{2} \)
1. **Force due to charge at \( -a \)**:
The distance from \( \frac{q}{2} \) to the charge at \( -a \) is \( a + x \).
The force \( F_1 \) exerted by the charge at \( -a \) on \( \frac{q}{2} \) is given by Coulomb's law:
\[ F_1 = k \frac{q \cdot \frac{q}{2}}{(a + x)^2} \]
This force is directed to the right (positive x-direction).

2. **Force due to charge at \( +a \)**:
The distance from \( \frac{q}{2} \) to the charge at \( +a \) is \( a - x \).
The force \( F_2 \) exerted by the charge at \( +a \) on \( \frac{q}{2} \) is:
\[ F_2 = k \frac{q \cdot \frac{q}{2}}{(a - x)^2} \]
This force is directed to the left (negative x-direction).

### Step 4: Net Force Calculation
The net force \( F \) acting on the charge \( \frac{q}{2} \) is:
\[ F = F_1 - F_2 = k \frac{q \cdot \frac{q}{2}}{(a + x)^2} - k \frac{q \cdot \frac{q}{2}}{(a - x)^2} \]

### Step 5: Simplifying the Net Force
To simplify, factor out the common terms:
\[ F = k \frac{q \cdot \frac{q}{2}}{(a + x)^2 (a - x)^2} \left( (a - x)^2 - (a + x)^2 \right) \]
Calculating \( (a - x)^2 - (a + x)^2 \):
\[ = (a^2 - 2ax + x^2) - (a^2 + 2ax + x^2) = -4ax \]
Thus, the net force becomes:
\[ F = -k \frac{q^2}{2} \cdot \frac{4ax}{(a + x)^2 (a - x)^2} \]

### Step 6: Analyzing the Motion
The net force \( F \) is proportional to \( -x \) (i.e., \( F \propto -x \)), indicating that the force acts in the opposite direction to the displacement. This is a characteristic of simple harmonic motion (SHM).

### Conclusion
Since the force acting on the charge \( \frac{q}{2} \) is proportional to the negative of its displacement from the equilibrium position (the origin), the charge will execute simple harmonic motion along the x-axis.

### Final Answer
The correct option is that the particle executes simple harmonic motion along the x-axis.

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